Skip to main content

Science Chemistry CHEM:ATOM FOC 2E CL (TEXT)

The percentage of a sample’s original radioactivity that remains after five half-lives. Solution: 3.125 % percent of a sample’s original radioactivity remains after five half-lives. Explanation: Concept: All radioactive decay processes follow first order kinetics. Each radioactive nuclide has a characteristic half-life ( t 1 2 ) , the time interval during which the quantity of radioactive particles decreases to half of its initial quantity. After every half-life, the number of radioactive atoms left in the given sample is one half of the number of the radioactive atoms that were present at the beginning of that half-life. Formula: N t N 0 = 0.5 t t 1 2 = ( 0.5 ) n n = t t 1 2 Given: Initial radioactivity of the original sample N 0 = 100 % Number of half-lives, n = 5 Calculations: Let’s consider that the initial radioactivity of the original sample is N 0 = 100 % The integrated rate law for the radioactive decays is N t N 0 = ( 0.5 ) n We can represent the number of half-lives, n . N t 100 = ( 0.5 ) 5 N t 100 = 0.03125 So % of the sample original radioactivity that remains after five half-lives is 0.03125 × 100 = 3.125 % So 3.125 % percent of a sample’s original radioactivity remains after five half-lives. Conclusion: The percentage of a sample’s original radioactivity remaining is calculated from its half life.

The percentage of a sample’s original radioactivity that remains after five half-lives. Solution: 3.125 % percent of a sample’s original radioactivity remains after five half-lives. Explanation: Concept: All radioactive decay processes follow first order kinetics. Each radioactive nuclide has a characteristic half-life ( t 1 2 ) , the time interval during which the quantity of radioactive particles decreases to half of its initial quantity. After every half-life, the number of radioactive atoms left in the given sample is one half of the number of the radioactive atoms that were present at the beginning of that half-life. Formula: N t N 0 = 0.5 t t 1 2 = ( 0.5 ) n n = t t 1 2 Given: Initial radioactivity of the original sample N 0 = 100 % Number of half-lives, n = 5 Calculations: Let’s consider that the initial radioactivity of the original sample is N 0 = 100 % The integrated rate law for the radioactive decays is N t N 0 = ( 0.5 ) n We can represent the number of half-lives, n . N t 100 = ( 0.5 ) 5 N t 100 = 0.03125 So % of the sample original radioactivity that remains after five half-lives is 0.03125 × 100 = 3.125 % So 3.125 % percent of a sample’s original radioactivity remains after five half-lives. Conclusion: The percentage of a sample’s original radioactivity remaining is calculated from its half life.

Solution Summary: The author explains that all radioactive decay processes follow first order kinetics. The number of atoms left in the sample is one half of the number that were present at the beginning of that half-life.

CHEM:ATOM FOC 2E CL (TEXT)

CHEM:ATOM FOC 2E CL (TEXT)

2nd Edition

ISBN: 9780393284218

Author: Stacey Lowery Bretz, Natalie Foster, Thomas R. Gilbert, Rein V. Kirss

Publisher: WW Norton & Co

See similar textbooks

Concept explainers

Writing and Balancing Nuclear Equations

A nuclear equation is a symbolic representation of a nuclear reaction using certain symbols. It is studied in nuclear chemistry and nuclear physics.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 21, Problem 21.33QA

Chapter 21, Problem 21.35QA

Chapter 21 Solutions

CHEM:ATOM FOC 2E CL (TEXT)

Ch. 21 - Prob. 21.1VP Ch. 21 - Prob. 21.2VP Ch. 21 - Prob. 21.3VP Ch. 21 - Prob. 21.4VP Ch. 21 - Prob. 21.5VP Ch. 21 - Prob. 21.6VP Ch. 21 - Prob. 21.7VP Ch. 21 - Prob. 21.8VP Ch. 21 - Prob. 21.9VP Ch. 21 - Prob. 21.10VP

Ch. 21 - Prob. 21.11VP Ch. 21 - Prob. 21.12VP Ch. 21 - Prob. 21.13QA Ch. 21 - Prob. 21.14QA Ch. 21 - Prob. 21.15QA Ch. 21 - Prob. 21.16QA Ch. 21 - Prob. 21.17QA Ch. 21 - Prob. 21.18QA Ch. 21 - Prob. 21.19QA Ch. 21 - Prob. 21.20QA Ch. 21 - Prob. 21.21QA Ch. 21 - Prob. 21.22QA Ch. 21 - Prob. 21.23QA Ch. 21 - Prob. 21.24QA Ch. 21 - Prob. 21.25QA Ch. 21 - Prob. 21.26QA Ch. 21 - Prob. 21.27QA Ch. 21 - Prob. 21.28QA Ch. 21 - Prob. 21.29QA Ch. 21 - Prob. 21.30QA Ch. 21 - Prob. 21.31QA Ch. 21 - Prob. 21.32QA Ch. 21 - Prob. 21.33QA Ch. 21 - Prob. 21.34QA Ch. 21 - Prob. 21.35QA Ch. 21 - Prob. 21.36QA Ch. 21 - Prob. 21.37QA Ch. 21 - Prob. 21.38QA Ch. 21 - Prob. 21.39QA Ch. 21 - Prob. 21.40QA Ch. 21 - Prob. 21.41QA Ch. 21 - Prob. 21.42QA Ch. 21 - Prob. 21.43QA Ch. 21 - Prob. 21.44QA Ch. 21 - Prob. 21.45QA Ch. 21 - Prob. 21.46QA Ch. 21 - Prob. 21.47QA Ch. 21 - Prob. 21.48QA Ch. 21 - Prob. 21.49QA Ch. 21 - Prob. 21.50QA Ch. 21 - Prob. 21.51QA Ch. 21 - Prob. 21.52QA Ch. 21 - Prob. 21.53QA Ch. 21 - Prob. 21.54QA Ch. 21 - Prob. 21.55QA Ch. 21 - Prob. 21.56QA Ch. 21 - Prob. 21.57QA Ch. 21 - Prob. 21.58QA Ch. 21 - Prob. 21.59QA Ch. 21 - Prob. 21.60QA Ch. 21 - Prob. 21.61QA Ch. 21 - Prob. 21.62QA Ch. 21 - Prob. 21.63QA Ch. 21 - Prob. 21.64QA Ch. 21 - Prob. 21.65QA Ch. 21 - Prob. 21.66QA Ch. 21 - Prob. 21.67QA Ch. 21 - Prob. 21.68QA Ch. 21 - Prob. 21.69QA Ch. 21 - Prob. 21.70QA Ch. 21 - Prob. 21.71QA Ch. 21 - Prob. 21.72QA Ch. 21 - Prob. 21.73QA Ch. 21 - Prob. 21.74QA Ch. 21 - Prob. 21.75QA Ch. 21 - Prob. 21.76QA Ch. 21 - Prob. 21.77QA Ch. 21 - Prob. 21.78QA Ch. 21 - Prob. 21.79QA Ch. 21 - Prob. 21.80QA Ch. 21 - Prob. 21.81QA Ch. 21 - Prob. 21.82QA Ch. 21 - Prob. 21.83QA Ch. 21 - Prob. 21.84QA Ch. 21 - Prob. 21.85QA Ch. 21 - Prob. 21.86QA Ch. 21 - Prob. 21.87QA Ch. 21 - Prob. 21.88QA Ch. 21 - Prob. 21.89QA Ch. 21 - Prob. 21.90QA Ch. 21 - Prob. 21.91QA Ch. 21 - Prob. 21.92QA Ch. 21 - Prob. 21.93QA Ch. 21 - Prob. 21.94QA Ch. 21 - Prob. 21.95QA Ch. 21 - Prob. 21.96QA Ch. 21 - Prob. 21.97QA Ch. 21 - Prob. 21.98QA Ch. 21 - Prob. 21.99QA Ch. 21 - Prob. 21.100QA Ch. 21 - Prob. 21.101QA Ch. 21 - Prob. 21.102QA Ch. 21 - Prob. 21.103QA Ch. 21 - Prob. 21.104QA Ch. 21 - Prob. 21.105QA Ch. 21 - Prob. 21.106QA Ch. 21 - Prob. 21.107QA Ch. 21 - Prob. 21.108QA Ch. 21 - Prob. 21.109QA Ch. 21 - Prob. 21.110QA Ch. 21 - Prob. 21.111QA Ch. 21 - Prob. 21.112QA Ch. 21 - Prob. 21.113QA Ch. 21 - Prob. 21.114QA Ch. 21 - Prob. 21.115QA Ch. 21 - Prob. 21.116QA Ch. 21 - Prob. 21.117QA Ch. 21 - Prob. 21.118QA Ch. 21 - Prob. 21.119QA Ch. 21 - Prob. 21.120QA Ch. 21 - Prob. 21.121QA Ch. 21 - Prob. 21.122QA Ch. 21 - Prob. 21.123QA Ch. 21 - Prob. 21.124QA Ch. 21 - Prob. 21.125QA Ch. 21 - Prob. 21.126QA Ch. 21 - Prob. 21.127QA Ch. 21 - Prob. 21.128QA Ch. 21 - Prob. 21.129QA Ch. 21 - Prob. 21.130QA

Knowledge Booster

Background pattern image

Chemistry

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99m is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave?
What are the two principal differences between nuclear reactions and ordinary chemical changes?

Recommended textbooks for you

Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry for Today: General, Organic, and Bioche...
Chemistry
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Cengage Learning
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning

Text book image

Chemistry: Matter and Change

Chemistry

ISBN:9780078746376

Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom

Publisher:Glencoe/McGraw-Hill School Pub Co

Text book image

Chemistry: The Molecular Science

Chemistry

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:Cengage Learning

Text book image

Chemistry for Today: General, Organic, and Bioche...

Chemistry

ISBN:9781305960060

Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Publisher:Cengage Learning

Text book image

General, Organic, and Biological Chemistry

Chemistry

ISBN:9781285853918

Author:H. Stephen Stoker

Publisher:Cengage Learning

Text book image

Chemistry: Principles and Reactions

Chemistry

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:Cengage Learning

Text book image

Introductory Chemistry: A Foundation

Chemistry

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Cengage Learning

SEE MORE TEXTBOOKS